With the same assumption as moritz's (that neither list has duplicate entries), you might try this: 

sub Intersection {
   my ($refA, $refB) = @_;
   my %B;
   @B{@$refB} = (1) x @$refB;  # hash the second list
   my $intersects;
   for(@$refA) {
      ++$intersects if exists $B{$_}  # 'exists' is fast!
   }
   return $intersects;
}

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In reply to Re: improving the speed by hbm
in thread improving the speed by perl_lover_always

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